Prove: $\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$, where $a^2 + b^2 = c^2$

Question

Could you help me proving this?

$$\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$$ where $c$ is the length of the hypotenuse of a right triangle, and $a$ and $b$ are the lengths of the other sides.

I tried that but not sure if it helps and what to do next:

$$\frac{\log\left(a\right)}{\log\left(c+b\right)}+\frac{\log\left(a\right)}{\log\left(c-b\right)}=\frac{2\log^2\left(a\right)}{\log\left(c+b\right)\log\left(c-b\right)}$$

Thank you in advance!

Answer 1

The change of base formula for logarithms says that $$ \log_xy=\frac{1}{\log_yx} $$ assuming $x$ and $y$ positive and different from $1$. Thus, assuming $a\ne1$, $c-b\ne1$ and $c+b\ne1$, we have \begin{align} \log_{c+b}a+\log_{c-b}a &= \frac{1}{\log_a(c+b)}+\frac{1}{\log_a(c-b)}\\[6px] &= \frac{\log_a(c-b)+\log_a(c+b)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px] &= \frac{\log_a\bigl((c-b)(c+b)\bigr)}{\log_a(c+b)\cdot\log_a(c-b)}\\[6px] &= \frac{\log_a(c^2-b^2)}{\log_a(c+b)\cdot\log_a(c-b)} \end{align} Since $c^2-b^2=a^2$, the numerator is $\log_aa^2=2$ and, applying again the change of base, $$ \log_{c+b}a+\log_{c-b}a=2\log_{c+b}a\cdot\log_{c-b}a $$

If $a=1$ the equality is obvious. If either $c+b=1$ or $c-b=1$, the equality doesn't make sense.

Answer 2

We have $$a^2+b^2=c^2$$ so that $$(c+b)(c-b)=c^2-b^2=a^2.$$ Take the logarithm of both sides $$\log(c+b)+\log(c-b)=2\log(a).$$

Divide by $\log(c+b)$:

$$\frac{\log(c+b)}{\log(c+b)}+\frac{\log(c-b)}{\log(c+b)}=2\frac{\log(a)}{\log(c+b)}$$

Simplify, using the fact that $\frac{\log(x)}{\log(y)}=\log_y(x)$:

$$1+\log_{c+b}(c-b)=2\log_{c+b}(a).$$

Multiply by $\log_{c-b}(a)$:

$$\log_{c-b}(a)+\log_{c-b}(a)\log_{c+b}(c-b)=2\log_{c+b}(a)\log_{c-b}(a).$$

Finally, $$\log_{c-b}(a)\log_{c+b}(c-b)=\frac{\log(a)}{\log(c-b)}\frac{\log(c-b)}{\log(c+b)}=\frac{\log(a)}{\log(c+b)}=\log_{c+b}(a).$$